NEET Physics Current Electricity MCQs Set 23

Answer: (a) \(-\frac{a}{2b}\)

Explanation:
1. \(E=αt+b{t}^{2}.\) At temperature of inversion \(E\) is minimum
2. \(i.e., E=0\)
3. \(∴α{t}_{i}+b{t}_{i}^{2}=0, i.e., {t}_{1}=-\frac{a}{b}\)

Answer: (c) 1 : 1

Explanation:
1. The specific resistance (\(ρ\)) is the characteristic of the material of conductor. Its value depends only on the material of conductor and its temperature. Its value does not depend on the length and area of cross-section of the conductor.

Answer: (c) Atomic mass/Valency

Explanation:
1. \(m=zq, z=\) atomic mass / valence

Answer: (b) \(2 Ω\)

Explanation:
1. In series, \({i}_{1}=\frac{2E}{2+2r}\)
2. In parallel, \({i}_{2}=\frac{E}{2+\frac{r}{2}}=\frac{2E}{4+r}\)
3. Since \({i}_{1}={i}_{2}⇒\frac{2E}{4+r}=\frac{2E}{2+2r}⇒r=2Ω\)

Answer: (c) 480 cm

Explanation:
1. At null point, no current flows through galvanometer. Therefore, the resistance connected in series with galvanometer, at null point position will be in effective. Thus, null point is obtained on potentiometer wire at the original position after the removal wire at the original position after the removal of series resistance from galvanometer

Answer: (d) \(\frac{{t}_{2}}{{t}_{1}}\)

Explanation:
1. Resistance of a conductor varies linearly with temperature as
2. \({R}_{t}={R}_{0}\left(1+αt\right)\)
3. For the first conductor
4. \({T}_{{t}_{1}}={R}_{0}\left(1+{α}_{1}{t}_{1}\right)\)
5. \(or {α}_{1}=\frac{R{t}_{1}-{R}_{0}}{t} ….\left(i\right)\)
6. Similarly, for second conductor
7. \({α}_{2}=\frac{R{t}_{2}-{R}_{0}}{{t}_{2}} ……..\left(ii\right)\)
8. From Eq. (i) and Eq. (ii), we get
9. \(\frac{{α}_{1}}{{α}_{2}}=\frac{{t}_{2}}{{t}_{1}}\)

Answer: (d) 96500 \(coulombs\)

Explanation:
1. 96500 \(coulombs\) of charge is needed to deposit one gram equivalent of an element at an electrode

Answer: (b) 2A

Explanation:
1. Current through external resistance,
2. \(i=\frac{nE}{nr+R}=\frac{5×2}{5×0.2+4}=2A\)

Answer: (d) 1 columb ×1 volt=1 watt

Explanation:
1. \(1 \mathrm{coulomb\ ×1volt=1\ joule}\)
2. Hence, option (d) is incorrect.

Answer: (c) Is excited

Explanation:
1. Human body, though has a large resistance of the order, of \(KΩ\left(\mathrm{say\ 10}kΩ\right),\) is very sensitive to minute currents even as low as a few \(mA.\) Electrons, excites and disorders the nervous system of the body and hence one fails to control the activity of the body

Answer: (a) \(\sqrt{{R}_{1}{R}_{2}}\)

Explanation:
1. Current given by cell
2. \(I=\frac{E}{R+r}\)
3. Power delivered in first case
4. \({P}_{1}={I}^{2}{R}_{1}={\left(\frac{E}{{R}_{1}+r}\right)}^{2}{R}_{1}\)
5. Power delivered in second case
6. \({P}_{2}={I}^{2}{R}_{2}={\left(\frac{E}{{R}_{2}+r}\right)}^{2}{R}_{2}\)
7. Power delivered is same in the both the cases.
8. \({\left(\frac{E}{{R}_{1}+r}\right)}^{2}{R}_{1}={\left(\frac{E}{{R}_{2}+r}\right)}^{2}{R}_{2}\)
9. \(\frac{{R}_{1}}{{\left({R}_{1}+r\right)}^{2}}=\frac{{R}_{2}}{{\left({R}_{2}+r\right)}^{2}}\)
10. \({R}_{1}\left({R}_{2}^{2}+{r}^{2}+2{R}_{2}r\right)={R}_{2}\left({R}_{1}^{2}+{r}^{2}+2{R}_{1}r\right)\)
11. \({R}_{1}{R}_{2}^{2}+{R}_{1}{r}^{2}+2{R}_{1}{R}_{2}r={R}_{2}{R}_{1}^{2}+{R}_{2}{r}^{2}+2{R}_{1}{R}_{2}r\)
12. \({R}_{1}{R}_{2}^{2}-{R}_{2}{R}_{1}^{2}={R}_{2}{r}^{2}-{R}_{1}{r}^{2}\)
13. \({R}_{1}{R}_{2}\left({R}_{2}-{R}_{1}\right)={r}^{2}\left({R}_{2}-{R}_{1}\right)\)
14. \(r=\sqrt{{R}_{1}{R}_{2}}\)

Answer: (c) 3.5 \(w\)g

Explanation:
1. \({m}_{\mathrm{Ag}}/{m}_{\mathrm{Zn}}\) = \({E}_{\mathrm{Ag}}\)/\({E}_{\mathrm{Zn}}\) = 108/31
2. \({\mathrm{or} m}_{\mathrm{Ag}}={m}_{\mathrm{Zn}}×108/31\)= \(w×108/31\)
3. 3.48 w = 3.5 w

Answer: (d) Antimony-bismuth

Explanation:
1. Thermocouples are widely used type of temperature sensor and are used as means to convert thermal potential difference into electric potential difference, using different combinations of metals. If metals were arranged according to their heating contrasts a series were formed antimony, iron, zinc, silver, gold, lead, mercury, copper, platinum and bismuth. The greater the heating contrasts between metals, the greater the electromotive force (EMF). Antimony and bismuth formed the best junction for emf.

Answer: (c) 0.126 g

Explanation:
1. According to Faraday’s law of electrolysis
2. \(m=zq\)
3. \(\frac{{m}_{\mathrm{zn}}}{{m}_{\mathrm{Cu}}}=\frac{{z}_{\mathrm{Zn}}}{{z}_{\mathrm{Cu}}}\)
4. \(∴\frac{0.13}{{m}_{\mathrm{Cu}}}=\frac{32.5}{31.5}\)
5. \({ m}_{\mathrm{Cu}}=\)0.126 g.

Answer: (c) Yes, if another identical battery is joined in opposition

Explanation:
1. If an identical battery is connected in opposition, net emf\(=E-E=0\)and the current through circuit will be zero, although each one of them has constant emf.

Answer: (a) Unchanged

Explanation:
1. Specific resistance is independent of dimensions of conductor but depends on nature of conductor.

Answer: (b) \(9.67×{10}^{4}C{g}_{\mathrm{eq}}^{-1}\)

Explanation:
1. Charge \(→q=\frac{m}{z}\)
2. Where, \(z=\) electro chemical equivalent
3. \(∴ z=\frac{m}{It}=\frac{0.067}{1×60}=0.001117\)
4. \(q=\frac{108}{0.001117}= 9.67×{10}^{4}C/{g}_{\mathrm{eq}}\)

Answer: (c) \(400℃\)

Explanation:
1. Let temperature of cold junction be \(0℃\) and that of hot junction be \(T℃.\) The relation for thermo-emf is given by
2. \(E=AT-\frac{1}{2} B{T}^{2}\)
3. Given, \(A=16, B=0.08\)
4. \(∴E=16T-\frac{1}{2}×0.08×{T}^{2}\)
5. Since, at temperature of inversion emf is zero, we have
6. \(0=16T-0.04 {T}^{2}\)
7. \(⇒T=\frac{16}{0.04}=400℃\)

Answer: (c) \(2.3 gm\)

Explanation:
1. Amount of metallic sodium appearing,
2. \(m=Zit=\left(\frac{A}{VF}\right)it=\left(\frac{23}{1×96500}\right)×16×10×60=2.3 gm\)

Answer: (c) \(975 Ω\) in series

Explanation:
1. \(R=\frac{V}{{i}_{g}}-G=\frac{6}{6×{10}^{-3}}-25=975Ω\) [In series]

Answer: (b) \(\frac{r}{10}\)

Explanation:
1. To obtain minimum resistance, all resistors must be connected in parallel.
2. Hence equivalent resistance of combination of combination \(=\frac{r}{10}\)

Answer: (c) Zero

Explanation:
1. In the absence of electric field (or potential difference applied) across the conductor, the average thermal velocity of electron is zero

Answer: (a) 3V

Explanation:
1. Given, \({E}_{1}=1.5V, {l}_{1}=27 cm,\)
2. \({l}_{2}=54 cm, {E}_{2}=?\)
3. \(\frac{{E}_{1}}{{E}_{2}}=\frac{{l}_{1}}{{l}_{2}}\)
4. \(\mathrm{or}{E}_{2}=\frac{{E}_{1}{l}_{2}}{{l}_{1}}\)
5. \(\mathrm{or}{E}_{2}=\frac{1.5×54}{27}\)
6. \({E}_{2}=3V\)

Answer: (a) Shifts towards right

Explanation:
1. As the resistance of metal increases on increasing the temperature, so resistance of metal conductor in left arm will increase on heating. For meter bridge
2. \(\frac{{R}_{1}}{{R}_{2}}=\frac{l}{100-l}\)
3. As \({R}_{1}\) increases \(l\) also increases.
4. Hence, balancing point shifts towards right.

Answer: (b) 3:1

Explanation:
1. As the current and the other factors are same for both the galvanometers
2. \(N ∝\mathrm{tan}\,θ\)
3. \(\frac{{N}_{1}}{{N}_{2}}=\frac{\mathrm{tan}\,60°}{\mathrm{tan}\,30°}=\frac{\sqrt{3}}{1/\sqrt{3}}\)
4. \(⟹\frac{{N}_{1}}{{N}_{2}}=3\)